Calculating Wavelength and Energy of Excited Hydrogen Atom | Help Guide

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SUMMARY

The discussion focuses on calculating the wavelength and energy of a photon absorbed by a hydrogen atom transitioning from the ground state (n=1) to the second excited state (n=3). The formula used is 1/λ = (1.097 x 107 m-1)(1/n12 - 1/n22), resulting in a wavelength of 9.72 x 10-8 m. The energy of the photon is calculated using E = hc/λ, yielding 2.045 x 10-18 J. The correct value for the second excited state is n=3, not n=4.

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fdizzy
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help please, for question: a hydrogen atom, initially in ground state, absorbs a photon, and is excited to its second excited state. What were the wavelength and the energy of the photon by the atom. Did i do it right.

1/λ = (1.097*10^7)(1/n1^2 - 1/n2^2)

1/λ = (1.097*10^7)(1/1^2 - 1/4^2)

λ = 9.72*10^-8 m


Energy = hc/λ

= (6.626*10^-34*3*10^8)/(9.72*10^-8)

= 2.045*10^-18 J
 
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You appear to be using 4 as the value of n for the second excited state. Why is that?
 
oh it should be 2 right ?
 
n = 1 is the ground state, which is not an excited state. So the first excited state would be...
 
so would it be 2
 
i think my n=1 would be 1 its grund state, excited state 2, would be n=3 ?
 
That's how I see it. :smile:
 

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